how to redirect multiline bash command into a variable

Question

I want to redirect this command to a variable:

{
    printf 'scan on\n\n'
    sleep 10
    printf 'quit \n\n'  
} | bluetoothctl

I tried to do this:

variable=$(printf 'scan on\n\n' sleep 10 printf 'quit \n\n' | bluetoothctl)
echo $variable

but I am getting blank output. I also tried to split it:

variable=$(printf 'scan on\n\n' | bluetoothctl)
sleep 10
printf 'quit \n\n'| bluetoothctl
echo $variable

but this time I only get

Agent registered
[bluetooth]# quit

How do I redirect the output of the first command which is printed to stdout when you run it, into a variable?

Answer 1

Just:

variable=$(
  {
    printf 'scan on\n\n'
    sleep 10
    printf 'quit \n\n'  
  } | bluetoothctl
)

The inside of command substitutions can be any shell code¹ and doesn't have to be on one line.

Here, you want to capture the output of the whole pipeline (in effect, that will be the output of bluetoothctl since all the other commands have their stdout redirected to the pipe that goes to bluetoothctl).

You could also capture the output of bluetoothctl only with:

{
  printf 'scan on\n\n'
  sleep 10
  printf 'quit \n\n'  
} | variable=$(bluetoothctl)

But in bash, for that to work, you'd need to set the lastpipe option (shopt -s lastpipe, only works for non-interactive shell instances), for that last pipe component not to be run in a subshell.

Here however, you don't need the pipeline, you can just do:

variable=$(bluetoothctl --timeout 10 scan on)

If you're going to display the contents of the variable, remember the syntax is:

printf '%s\n' "$variable"

not echo $variable.

More details on that at:

• Why is printf better than echo?
• When is double-quoting necessary?

Here the output of bluetoothctl contains colouring escape sequences as well as escape sequences used to clear the contents of lines. If you leave $variable unquoted, its contents will be subject to split+glob resulting in a random list of words to pass to echo which echo will output space-separated. So you'll end up with everything on one line and if the last word ends in a clear-line sequence (which it does in my tests: \r\e[K being Carriage-Return followed by the Clear-To-End-Of-Line sequence), you'll just see a blank line.

---

^{¹ though some shells including bash used to choke on code with unmatched parenthesis like when the code included case statements like case $x in foo) or comments with unbalanced ) which the shell parser was mistaking for the ) closing the command substitution.}

Answer 2

The mistake is the lack of semicolons (that are otherwise equivalent to newlines). And if you want to see the structure of the output, add double-quotes. Try:

variable=$( { printf 'scan on\n\n' ; sleep 10 ; printf 'quit \n\n' ; } | bluetoothctl)
echo "$variable"

Answer 3

A possible alternative could be a simple expect script:

#!/usr/bin/expect
spawn bluetoothctl
send "scan on\r"
sleep 10
send "quit \r"
expect eof

\r equals "Return / pressing enter". Make executable and run.